Solve for $x$ and $y$ using substitution. ${6x+6y = 12}$ ${x = -2y-1}$
Explanation: Since $x$ has already been solved for, substitute $-2y-1$ for $x$ in the first equation. ${6}{(-2y-1)}{+ 6y = 12}$ Simplify and solve for $y$ $-12y-6 + 6y = 12$ $-6y-6 = 12$ $-6y-6{+6} = 12{+6}$ $-6y = 18$ $\dfrac{-6y}{{-6}} = \dfrac{18}{{-6}}$ ${y = -3}$ Now that you know ${y = -3}$ , plug it back into $\thinspace {x = -2y-1}\thinspace$ to find $x$ ${x = -2}{(-3)}{ - 1}$ $x = 6 - 1$ ${x = 5}$ You can also plug ${y = -3}$ into $\thinspace {6x+6y = 12}\thinspace$ and get the same answer for $x$ : ${6x + 6}{(-3)}{= 12}$ ${x = 5}$